种子填充找连通块 floodfill

Description

Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors. 

Given a diagram of Farmer John's field, determine how many ponds he has.

Input

* Line 1: Two space-separated integers: N and M 

* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.

Output

* Line 1: The number of ponds in Farmer John's field.

Sample Input

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

Sample Output

3

Hint

OUTPUT DETAILS: 

There are three ponds: one in the upper left, one in the lower left,and one along the right side.

 

题目意思：输入行数列数，m，n。然后输入像上面的字符。找连着的W有多少块.....

解题思路：首先用一个二维字符数组把输入的存起来....  然后一个一个的循环，如果是W并且还没有被标记过就进入zhao这个函数

zhao函数：找嘛，大概意思就是，看当前位置的8个方向有没有连通的，这里用到了递归。希望代码上的注释，对你有帮助。

（其实，自己对这代码的意思也是似懂非懂，找书打出来的）

代码如下：   

 

#include <iostream>
#include <cstring>
using namespace std;
const int maxn=1000;
int m,n;    //这里将m，n定义在主函数外，作为全局变量，好被zhao函数调用
char s[maxn][maxn];
int d[maxn][maxn];
void zhao(int r,int c,int b)  
{
    if(r<0||r>=m||c<0||c>=n)    //出界的，不要
        return;
    if(d[r][c]>0||s[r][c]!='W')  // 不是W，或者已经访问标记过了的格子
        return;
    d[r][c]=b;   //给访问过的格子标记
    for(int dr=-1;dr<=1;dr++)   //这两个循环是表示一个格子的八个方向
    {
        for(int dc=-1;dc<=1;dc++)
            if(dr!=0||dc!=0)  //这里不要0，0的格子，因为这就是它本身，并没有动
                zhao(r+dr,c+dc,b);  //递归，将dr,dc加上去，这样就寻找了附近的格子
    }
}

int main()
{

    memset(d,0,sizeof(d));   //将d数组清零，好标记
    while(cin>>m>>n)
    {
        int flag=0;
        for(int i=0; i<m; i++)
        {
            for(int j=0; j<n; j++)
                cin>>s[i][j];
        }
        for(int i=0; i<m; i++)
        {
            for(int j=0; j<n; j++)
            {
                if(d[i][j]==0&&s[i][j]=='W')
                    zhao(i,j,++flag);
            }
        }
        cout<<flag<<endl;
    }
    return 0;
}